In the hydrogen spectrum, λ be the wavelength of first transition line of Lyman series. The wavelength difference will be "aλ'' between the wavelength of 3ʳᵈ transition line of Paschen series and that of 2ⁿᵈ transition line of Balmer series where a= ___________.

Solution

Physics sir · 3 steps

Write all three wavelengths as multiples of 1/R, then the ratio falls out with no constants left.

1. Rydberg formula. For a jump from level n₂ down to level n₁, use the formula below. The k-th line of a series starts from n₂ = n₁ + k. So the Lyman first line is 2 → 1, the Balmer second line is 4 → 2, and the Paschen third line is 6 → 3.

   1/λ = R(1/n₁² - 1/n₂²)

   Line	Transition	1/λ	Wavelength
   Lyman, 1st	2 → 1	R(1 - 1/4) = 3R/4	λ = 4/3R
   Balmer, 2nd	4 → 2	R(1/4 - 1/16) = 3R/16	16/3R
   Paschen, 3rd	6 → 3	R(1/9 - 1/36) = R/12	12/R = 36/3R

2. Take the difference. 36/3R - 16/3R = 20/3R. Since λ = 4/3R, this equals 5 × 4/3R = 5λ.
3. So. The wavelength difference is 5λ, so a = 5.

The idea to keep. The k-th line of a series starts from n₁ + k. Keep every wavelength as a fraction times 1/R and put them over a common denominator before you compare them.

Answer from the NTA final answer key. Solution by PrepWiser.